Question: Let $f(x)=\dfrac{1}{x^2}+\dfrac{1}{x}$. $f'(1)=$
Answer: The strategy We can first rewrite each rational term of $f$ as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Once we have $f'(x)$, we can plug $x=1$ into it to find $f'(1)$. Rewriting rational terms as negative powers $\begin{aligned} f(x)&=\dfrac{1}{x^2}+\dfrac{1}{x} \\\\ &=x^{-2}+x^{-1} \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}(x^{-2}+x^{-1}) \\\\ &=\dfrac{d}{dx}(x^{-2})+\dfrac{d}{dx}(x^{-1}) \\\\ &=-2x^{-3}+(-1)x^{-2} \\\\ &=-\dfrac{2}{x^3}-\dfrac{1}{x^2} \end{aligned}$ Evaluating $f'(x)$ So we found that $f'(x)=-\dfrac{2}{x^3}-\dfrac{1}{x^2}$. Let's plug $x=1$ into $f'$ : $\begin{aligned} &\phantom{=}f'(1) \\\\ &=-\dfrac{2}{1^3}-\dfrac{1}{1^2} \\\\ &=-2-1 \\\\ &=-3 \end{aligned}$ In conclusion, $f'(1)=-3$.